\(\left|x^2-x+2\right|-3x-7=0\)
\(\Leftrightarrow\left|x^2-x+2\right|=3x+7\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+2=3x+7\\x^2-x+2=-3x-7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x-3x+2-7=0\\x^2-x+3x+2+7=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x-5=0\\x^2+2x+9=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-5\right)\left(x+1\right)=0\\x^2+2x+9=0\left(vonghiem\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
Vậy........................
Cách khác :
\(\text{|}x^2-x+2\text{|}-3x-7=0\)
⇔ \(\text{|}x^2-x+2\text{|}=3x+7\)
Do : \(\text{|}x^2-x+2\text{|}\) ≥ 0 ∀x
⇒ \(3x+7\) ≥ 0 ⇔ \(x\) ≥ \(-\dfrac{7}{3}\)
Bình phương hai vế của phương trình , ta được :
\(\left(x^2-x+2\right)=\left(3x+7\right)^2\)
⇔ \(\left(x^2-x+2\right)-\left(3x+7\right)^2=0\)
⇔ \(\left(x^2-x+2-3x-7\right)\left(x^2-x+2+3x+7\right)=0\)
⇔ \(\left(x^2-4x-5\right)\left(x^2+2x+9\right)=0\)
Do : \(x^2+2x+9=x^2+2x+1+8=\left(x+1\right)^2+8>0\)
⇒ \(x^2-4x+5=0\)
⇔ \(\left(x+1\right)\left(x-5\right)=0\)
⇔ \(x=-1\left(TM\right);x=5\left(TM\right)\)
KL...................
